Math, asked by ashokramchandani1972, 10 months ago

The value of k if kx^2 + 2(k+1)x+(3k+1)=0 has real and equal roots

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Answered by prakharagrawal0013

1

Answer:

Step-by-step explanation:

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Answered by sahilsuman933

2

Answer:

D= 0 (Condition For Equal Roots)

4(k+1)^2-4k(3k+1) = 0

4[k2+1+2k]-12k2-4k=0

4k2+4+8k-12k2-4k=0

-8k2+4k+4=0

2k2-k-k=0

2k2-2k+k-1=0

2k(k-1)+1(k-1)=0

(2k+1)(k-1)=0

k = -1/2 or 1

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